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USPshnik [31]
2 years ago
15

Pls pls pls answer quick

Mathematics
1 answer:
antoniya [11.8K]2 years ago
5 0

Answer:

5x+3=3x+15

Subtract 3 from both sides

5x=3x+12

Subtract 3x from both sides

2x=12

Divide both sides by 2

Step-by-step explanation:

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Lucy's school is celebrating easter by having a race in which the winner will get a basket of chocolate eggs. lucy is the fiftie
Alja [10]
The question also states that Lucy has a winning probability of 1/50, which means that she has 1 chance of winning if the total runners were 50. Therefore, there are 49 runners who may be faster than her.

The fact that Lucy is the 50th slowest runner, means that starting from the slowest she is in 50th position, therefore there are 49 runners that are slower than her.

The total number of runners will be the sum of those faster than Lucy, those slower than Lucy and Lucy:
49 + 49 + 1 = 99

There are 99 runners in Lucy's school.
7 0
3 years ago
CAN SOMEONE PLS HELP 6TH GRADE MATH ILL GIVE BRAINLIEST
ASHA 777 [7]

Answer:

14/100 and 7/50

Step-by-step explanation:

Basically you convert the decimal into a fraction, 14 percent is the same as 14/100 and if you simplify that then you get 7/50

4 0
2 years ago
Simply the expression <br> 6(6+6+2v) HELP PLEASE ( I know my computer screen is dirty lol)
olya-2409 [2.1K]

Answer:

12v + 62

Step-by-step explanation:

distribute the 6 to 6, 6, and 2v.

36 + 36 + 12v=

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8 0
3 years ago
Does x=5 satisfy 3(x+1)=3x+3
Svetradugi [14.3K]

Answer:

Yes

Step-by-step explanation:

Well let's solve it!

In order to solve this, plug in x into the equation.

Let's do the side on the LEFT of the equal sign first.

3(3+1) = 3(4) = 12

Now Let's do the side on the RIGHT of the equal sign.

3(3) + 3 = 9+3 = 12

So this means they are equal or congruent!


Hope this Helped!

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%
xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }

Let a = 1 and b the cosine product, and write them as

\dfrac{a - b}{x^2}

with

b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}

Now we use the identity

a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)

to rationalize the numerator. This gives

\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}

For the remaining limit, use the Taylor expansion for cos(x) :

\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)

where \mathcal{O}(x^4) essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)

\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0
2 years ago
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